Solve $(x+1)^y-x^z=1$ I want to know if there is another way to solve the equation:
$$(x+1)^y-x^z=1$$
for positive integers $x,y,z$ greater than $1$ without using the Catalan identity. Does exist? By now I only know that $x\mid y$.
 A: $(x+1)^y=x^z+1^z$ with $x,y,z\ge 2$. 
By Zsigmondy's theorem, unless $(x,y,z)=(2,2,3)$, $x^z+1^z$ has a prime divisor that does not divide $x+1$, which is impossible.
A: In Which is greater, 9899 or 9998? it was stablished that $(x+1)^x<x^{x+1}$ for all real $x>2.293$. On the other hand, $(x,y,z)=(1,1,1)$ ,$(2,2,3)$ and $(1,1,z)$ with $z\ge1$ are clearly solutions.
We have  $$(x+1)^y-x^z=1 \Rightarrow (x+1)^y>x^z\Rightarrow y<z$$ From
$$\begin{cases}(x+1)^x<x^{x+1}\text{ for all x>2.293}\\(x+1)^y>x^z\\y<z\end{cases}$$ we deduce easily $y<2.293$ then $y=1$ or $2$ which gives the only solutions above.
A: Two obvious remarks : first, we have a whole bunch of solutions given by $y=z=1$ (since $(x+1)-x = 1$), and there is a solution with $y \neq z$, namely $(x,y,z) = (2,2,3)$ since $(2+1)^2 - 2^3 = 9-8 = 1$, which you can find by inspection (I just started putting low exponents and figured it out by guessing). 
Second, a solution must satisfy $y \le z$, for if $y > z$, then since $x \ge 1$, 
$$
(x+1)^{y+1} = (x+1)^y (x+1) \ge (x+1)^y + 1 > x^z+1. 
$$
(Which makes sense, the bigger number should have a smaller exponent so that their difference can be small, e.g. $1$.)
Added : third remark, if $y=z > 1$, we have no solutions by the $a^n-b^n$ identity, for if $x \ge 1$, we have 
$$
(x+1)^z - x^z = ((x+1)-x)\left( \sum_{i=0}^{z-1} x^i (x+1)^{z-1-i} \right) \ge x^{z-1} + (x+1)^{z-1} > 1. 
$$
(This wasn't meant to be a full answer, just too big for a comment.)
Hope that helps,
